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3 years ago

How to get the greatest common of 45 and 81

2 answers:

5 0

Find the factors and get the ones common to both of them

45=3 times 3 times 5
81=3 times 3 times 3 times 3
so the factors common to both are 3 times 3 which is 9

kirill [66]3 years ago

3 0

You should draw a factor tree. It will help you a lot .

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The question.

Tanya [424]

The solution to this problem can be found by taking 20% of $62.50:

0.2(62.5) = 12.5

The way the question is worded is confusing, but I think it's asking how much of a discount you would get. You would receive a $12.50 discount.

3 0

3 years ago

Solve (3x – 2)= 49 using the Square Root Property (completing the square).

morpeh [17]

Answer:

(3x-2)=49

(3x-2)=7×7

3x=7×7+2

3x=49+2

3x=51

3x/3=51/3

×=17

3 0

2 years ago

g Suppose a factory production line uses 3 machines, A, B, and C for making bolts. The total output from the line is distributed

Vladimir [108]

Answer:

The probability that it came from A, given that is defective is 0.362.

Step-by-step explanation:

Define the events:

A: The element comes from A.

B: The element comes from B.

C: The element comes from C.

D: The elemens is defective.

We are given that P(A) = 0.25, P(B) = 0.35, P(C) = 0.4. Recall that since the element comes from only one of the machines, if an element is defective, it comes either from A, B or C. Using the probability axioms, we can calculate that

$P(D) = P(A\cap D) + P(B\cap D) + P(C\cap D)$

Recall that given events E,F the conditional probability of E given F is defined as

$P(E|F) = \frac{P(E\cap F)}{P(F)}$ , from where we deduce that

$P(E\cap F) = P(E|F)P(F)$ .

We are given that given that the element is from A, the probability of being defective is 5%. That is P(D|A) =0.05. Using the previous analysis we get that

$P(D) = P(A)P(D|A)+P(B) P(D|B) + P(C)P(D|C) = 0.05\cdot 0.25+0.04\cdot 0.35+0.02\cdot 0.4 = 0.0345$

We are told to calculate P(A|D), then using the formulas we have

$P(A|D) = \frac{P(A\cap D)}{P(D)}= \frac{P(D|A)P(A)}{P(D)}= \frac{0.05\cdot 0.25}{0.0345}= 0.36231884$

3 0

2 years ago

Find the equation of the line

kotegsom [21]

Of the line of what?

7 0

2 years ago

Read 2 more answers

each paperclip is 7/8 of an inch long and costs $.03. exactly enough paper clips are laid end to end yo have a total length of 5

GalinKa [24]

1 clipe..............7/8
x.......................56

7x/8 = 56
7x = 56*8
x = 56*8/7
x = 8*8
x = 64 clipes

1 clipe = 0,03
64 clipes = 0,03*64
64 clipes = 1,92

5 0

2 years ago